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Assume $G$,$\hat{G}$ are both free group of rank $n$,and $H$,$\hat{H}$ be their subgroups of index $k$ respectively,$h:H \rightarrow G$, $\hat{h}:\hat{H} \rightarrow\hat{G}$, are two epimorphisms. We call the triple $(H,G,h)$ and $(\hat{H},\hat{G},\hat{h})$ are equivalent if there is a isomorphism $\alpha:G \rightarrow \hat{G}$ satisfies (a)$\alpha(G)=\hat{G}$,$\alpha(H)=\hat{H}$ (b)$\alpha \circ h(x)=\hat{h} \circ \alpha(x)$,$\forall x \in H$ (1)Then I have three questions:(1)when are $(H,G,h)$ and $(\hat{H},\hat{G},\hat{h})$ equivalent?

(2)If we already know $(H,G,h)$ and $(\hat{H},\hat{G},\hat{h})$ are equivalent, how can we find an $\alpha$ ?

(3)For given $n$ and $k$,Can we give a classfy of all those triples in this way?

For all the questions above, I have considered simple the case $n=3$ and founded it's really difficulty for me. I don't know whether this question is trivial or not in Group Theory. If you can give me some useful advices i will be gratiful. Thanks!

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    $\begingroup$ There is no need to call anyone a genius. You should remove that from your question, as it does not add anything of value and, quite the contrary, probably detracts from it. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Oct 31, 2013 at 3:35
  • $\begingroup$ I think you need to fix the condition on your isomorphism $\alpha$. First, $h(x) \in \hat{H}$, so you can't apply $\alpha$ to it. Second, $\alpha(x) \in \hat{G}$, so you can't apply $\hat{h}$ to it. $\endgroup$
    – S. Carnahan ♦
    Commented Oct 31, 2013 at 13:54
  • $\begingroup$ @S.Carnahan: $\alpha :G \rightarrow \hat{G}$ is a isomorphic.$\alpha(G)=\hat{G}$. And $\alpha\ restrict\ on\ H\ is\ a\ isomorphic\ from\ H\ to\ \hat{H}$. So we can apply $\hat{h}$ to it. I'm sorry for did make it clear before. Thanks very much! $\endgroup$
    – Grub
    Commented Oct 31, 2013 at 14:13
  • $\begingroup$ The question seems too unnatural to be 'trivial'. Could you give some context? Where does it come from? $\endgroup$
    – HJRW
    Commented Oct 31, 2013 at 15:59
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    $\begingroup$ @HJRW such epimorphisms are called "virtual endomorphism" in the book "self-similar group". It's natural to give a classify of such object. $\endgroup$
    – Grub
    Commented Nov 1, 2013 at 2:04

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Here's a naive procedure to answer part (2) of your question. Let $S=(s_1,\ldots,s_n)$ be a minimal generating set for $H$. Using Whitehead's algorithm, say, we can enumerate minimal generating sets $(\hat{s}_i)$ of $\hat{H}$. Now the assignment $s_i\mapsto\hat{s}_i$ gives you a candidate automorphism $\alpha$.

Since $H$ is of finite index, we can check membership of $H$ and $\alpha(H)$, and hence determine whether or not $\alpha(H)=H$. To check that $\alpha\circ h=\hat{h}\circ\alpha$, it suffices to check thison the generators of $H$. This is a finite number of identities, which can be checked using the solution to the word problem in free groups.

By the way, I don't understand the difference between your Questions (1) and (3).

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  • $\begingroup$ Thanks for your answer! But I think it really not easy to find the solution to the word problem in free groups. In the case $n=3,k=2$ , this need to solve a equation of the following type:$h^{\circ 3} \dots=1$ ,Then I don't know how to deal with it.By question (3) I means that "what is the space of all those triples module the equivalence" $\endgroup$
    – Grub
    Commented Nov 2, 2013 at 10:50
  • $\begingroup$ @Grub - the word problem in free groups is trivial. Just write your elements as reduced words and you're done. Look at Lyndon and Schupp's book, for instance, for details. $\endgroup$
    – HJRW
    Commented Nov 2, 2013 at 12:14

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